How to find/construct functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$












0












$begingroup$


Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.



If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$



After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?



I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.






real-analysis number-theory







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$endgroup$








  • 1




    $begingroup$
    take a function f(n) arbitrarily and put g(n)=-f(n)e.
    $endgroup$
    – Minz
    Dec 12 '18 at 4:16












  • $begingroup$
    Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
    $endgroup$
    – GovernmentFX
    Dec 12 '18 at 4:17












  • $begingroup$
    @GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
    $endgroup$
    – Pinteco
    Dec 12 '18 at 4:24










  • $begingroup$
    Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 5:36
















0












$begingroup$


Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.



If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$



After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?



I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.






real-analysis number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    take a function f(n) arbitrarily and put g(n)=-f(n)e.
    $endgroup$
    – Minz
    Dec 12 '18 at 4:16












  • $begingroup$
    Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
    $endgroup$
    – GovernmentFX
    Dec 12 '18 at 4:17












  • $begingroup$
    @GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
    $endgroup$
    – Pinteco
    Dec 12 '18 at 4:24










  • $begingroup$
    Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 5:36














0












0








0





$begingroup$


Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.



If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$



After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?



I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.






real-analysis number-theory







share|cite|improve this question











$endgroup$




Here, all $f(n)$ and $g(n)$ mentioned, are functions that assume integers values for some integer $n>0$.



If $n>0$ and integer, we have that $I_n = int_0^1 x^ne^xdx = f(n)e+g(n)$ where $f(n)=(-1)^n!n$ and $g(n) = (-1)^{n+1}n!$ and $lim_{nto infty} f(n)e + g(n) = 0$. To find $f$ and $g$ above we can evaluate $I_0, I_1,I_2,...$ until a pattern is finded. And the limit is here $lim_{n to infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 : ?$



After that I wonder if there are other functions $f(n)$ and $g(n)$ such that $lim_{nto infty} f(n)e + g(n) = 0:?$ How to find they? What is the process of creating they?



I'm asking this because I wanted to do something similar for the catalan constant, that is, finding functions $f(n),g(n)$ such that $lim_{ntoinfty}f(n)(C-1)+g(n)C = 0$.






real-analysis number-theory




real-analysis number-theory






share|cite|improve this question















share|cite|improve this question













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edited Dec 12 '18 at 4:26







Pinteco

















asked Dec 12 '18 at 4:04









PintecoPinteco

731313




731313








  • 1




    $begingroup$
    take a function f(n) arbitrarily and put g(n)=-f(n)e.
    $endgroup$
    – Minz
    Dec 12 '18 at 4:16












  • $begingroup$
    Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
    $endgroup$
    – GovernmentFX
    Dec 12 '18 at 4:17












  • $begingroup$
    @GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
    $endgroup$
    – Pinteco
    Dec 12 '18 at 4:24










  • $begingroup$
    Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 5:36














  • 1




    $begingroup$
    take a function f(n) arbitrarily and put g(n)=-f(n)e.
    $endgroup$
    – Minz
    Dec 12 '18 at 4:16












  • $begingroup$
    Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
    $endgroup$
    – GovernmentFX
    Dec 12 '18 at 4:17












  • $begingroup$
    @GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
    $endgroup$
    – Pinteco
    Dec 12 '18 at 4:24










  • $begingroup$
    Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 12 '18 at 5:36








1




1




$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16






$begingroup$
take a function f(n) arbitrarily and put g(n)=-f(n)e.
$endgroup$
– Minz
Dec 12 '18 at 4:16














$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17






$begingroup$
Are there some condition(s) that $f,g$ must satisfy? Because if not then if you let $f,g$ be sequences with limit zero at infinity, your required limit will hold trivially.
$endgroup$
– GovernmentFX
Dec 12 '18 at 4:17














$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24




$begingroup$
@GovernmentFX $f,g$ must be always integers, for some positive integer $n$.
$endgroup$
– Pinteco
Dec 12 '18 at 4:24












$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36




$begingroup$
Since $e$ is irrational ${ne+m:n,m in mathbb Z}$ is dense in $mathbb R}$.
$endgroup$
– Kavi Rama Murthy
Dec 12 '18 at 5:36










1 Answer
1






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oldest

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0












$begingroup$

Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.



consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .



Let's define $f$ as follows.



$$ f(1) stackrel{df}{=} 1 $$
$$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$



So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.



And then we define $g$ to be the floor of $e cdot f(n)$ .



$$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$



Here are the first few values of $f(n)$



n    f(n)    exp(f(n))

1       1      2.7182817

2       2      5.4365635

3       3      8.154845

4       7     19.027973

5      39    106.012985





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    0












    $begingroup$

    Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.



    consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .



    Let's define $f$ as follows.



    $$ f(1) stackrel{df}{=} 1 $$
    $$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$



    So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.



    And then we define $g$ to be the floor of $e cdot f(n)$ .



    $$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$



    Here are the first few values of $f(n)$



    n    f(n)    exp(f(n))
    
1       1      2.7182817
    
2       2      5.4365635
    
3       3      8.154845
    
4       7     19.027973
    
5      39    106.012985





    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.



      consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .



      Let's define $f$ as follows.



      $$ f(1) stackrel{df}{=} 1 $$
      $$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$



      So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.



      And then we define $g$ to be the floor of $e cdot f(n)$ .



      $$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$



      Here are the first few values of $f(n)$



      n    f(n)    exp(f(n))
      
1       1      2.7182817
      
2       2      5.4365635
      
3       3      8.154845
      
4       7     19.027973
      
5      39    106.012985





      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.



        consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .



        Let's define $f$ as follows.



        $$ f(1) stackrel{df}{=} 1 $$
        $$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$



        So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.



        And then we define $g$ to be the floor of $e cdot f(n)$ .



        $$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$



        Here are the first few values of $f(n)$



        n    f(n)    exp(f(n))
        
1       1      2.7182817
        
2       2      5.4365635
        
3       3      8.154845
        
4       7     19.027973
        
5      39    106.012985





        share|cite|improve this answer











        $endgroup$



        Here's one way to do it. It's well defined but doesn't give you a way to compute $f(n)$ or $g(n)$ very efficiently.



        consider the fractional part of $e cdot n$, here written $mathrm{F}(e cdot n)$ .



        Let's define $f$ as follows.



        $$ f(1) stackrel{df}{=} 1 $$
        $$ f(n) stackrel{df}{=} min bigg{ k ;|; k in mathbb{N} ;land; k gt f(n-1) ;land; mathrm{F}(e cdot k) lt mathrm{F}(e cdot f(n-1)) bigg} $$



        So, each successive value in the sequence $f(n)$ is sent to the smallest natural number possible such that $mathrm{F}(,f(n) cdot e,)$ is strictly monotonically decreasing. The extra condition $k gt f(n-1)$ is, strictly speaking, redundant.



        And then we define $g$ to be the floor of $e cdot f(n)$ .



        $$ g(n) stackrel{df}{=} leftlfloor e cdot f(n) rightrfloor $$



        Here are the first few values of $f(n)$



        n    f(n)    exp(f(n))
        
1       1      2.7182817
        
2       2      5.4365635
        
3       3      8.154845
        
4       7     19.027973
        
5      39    106.012985






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 5:29

























        answered Dec 12 '18 at 5:21









        Gregory NisbetGregory Nisbet

        573312




        573312






























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